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Ex 9.4, 1 For each of the differential equations in Exercises 1 to 10, find the general solution : 𝑑𝑦/𝑑π‘₯=(1 βˆ’ cos⁑π‘₯)/(1 + cos⁑π‘₯ ) 𝑑𝑦/𝑑π‘₯ = (1 βˆ’ cos⁑π‘₯)/(1 + cos⁑π‘₯ ) We know that cos 2x = 2cos2 x βˆ’ 1 Putting x = π‘₯/2 cos 2π‘₯/2 = 2 cos2 π‘₯/2 βˆ’ 1 cos x = 2 cos2 π‘₯/2 βˆ’ 1 1 + cos x = 2cos2 π‘₯/2 We know cos 2x = 1 βˆ’ 2sin2 x Putting x = π‘₯/2 cos2 2π‘₯/2 = 1 βˆ’ 2 sin2 π‘₯/2 cos x = 1 βˆ’ 2sin2 π‘₯/2 1 βˆ’ cos x = 2sin2 π‘₯/2 Putting values in equation π’…π’š/𝒅𝒙 = (𝟐 〖〖𝐬𝐒𝐧〗^𝟐 〗⁑〖𝒙/πŸγ€— )/(𝟐 γ€–γ€–πœπ¨π¬γ€—^𝟐 〗⁑〖𝒙/πŸγ€— ) 𝑑𝑦 = (γ€–sin^2 〗⁑〖π‘₯/2γ€— )/cos^(2 )⁑〖π‘₯/2γ€— 𝑑π‘₯ 𝑑𝑦 = tan2 π‘₯/2 𝑑π‘₯ Putting tan2 π‘₯/2 = sec2 π‘₯/2 βˆ’ 1 π’…π’š = ("sec2 " 𝒙/𝟐 " βˆ’ 1" )𝒅𝒙 Integrating both sides We know that tan2 x + 1 = sec2 x tan2 x = sec2 x βˆ’ 1 Putting x = π‘₯/2 tan2 π‘₯/2 = sec2 π‘₯/2 βˆ’ 1 ∫1β–’π’…π’š = ∫1β–’(π’”π’†π’„πŸ 𝒙/πŸβˆ’πŸ) 𝒅𝒙 y = ∫1▒〖𝑠𝑒𝑐2 π‘₯/2 𝑑π‘₯βˆ’ ∫1▒𝑑π‘₯γ€— y = 𝟏/(𝟏/𝟐) tan 𝒙/𝟐 βˆ’ x + C y = 𝟐 tan 𝒙/𝟐 βˆ’ x + C is the general solution (As ∫1▒〖𝑠𝑒𝑐2 (π‘Žπ‘₯) 𝑑π‘₯γ€— = 1/π‘Ž tan ax + C)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.